PE Mechanical HVAC and Refrigeration Exam Prep
Start your PE Mechanical HVAC and Refrigeration exam prep with this original, unofficial problem, written to the October 2025 NCEES specification. All 36 problems, worked solutions, and the 12-week plan are free on this page.
Problem 1 · Sensible cooling load at sea level
HLP-01 · HVAC Loads and Psychrometrics, 1B · Select one
A space is supplied with 2,000 cfm of standard air at 55°F and is held at 75°F. Assume sea level conditions and steady state. What sensible cooling load does the supply air remove from the space?
Show answer and solution
Answer and solution
Answer: C. 43,200 Btu/h.
Qs = 1.08 × cfm × ΔT = 1.08 × 2,000 × (75 − 55) = 43,200 Btu/h. The 1.08 is 60 min/h × 0.075 lb/ft³ × 0.24 Btu/(lb·°F), so it holds only for standard, sea-level air. A leaves out the 20°F temperature difference. B is the answer at 5,000 ft (Problem 2). The exam assumes sea level unless a problem says otherwise. D uses the 4.5 total-heat factor, which multiplies an enthalpy change, not a temperature change.
Try a change: Move the same space to 5,000 ft. Each cubic foot of air now carries less mass, so the same 2,000 cfm removes less heat. Problem 2 works it out.
Source: NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1B.
By Castleport Test Prep Editorial Team · Exam facts last verified October 8, 2026
Topic weights · Studying from an older book? · All 36 problems · 12-week plan
The exam at a glance
- 80 questions, computer-based, offered year-round at Pearson VUE test centers
- 8 hours of exam time inside a 9-hour appointment (2-minute nondisclosure agreement, 8-minute tutorial, 50-minute scheduled break)
- $400 exam fee paid to NCEES; your licensing board may charge its own application fee
- The only reference on screen is the NCEES PE Mechanical Reference Handbook. No ASHRAE, SMACNA, or code books are supplied, and you can't bring your own.
- U.S. customary units only, and sea level conditions unless a problem says otherwise
- Multiple choice plus alternative item types (select-all, drag-and-drop, point-and-click, fill-in-the-blank), all scored right or wrong
- Pass/fail results, typically 7–10 days after the exam
- First-time pass rate: 71% of 799 examinees; repeat takers: 40% of 229 (NCEES table updated July 2026)
Sources: NCEES PE Mechanical exam page; HVAC and Refrigeration specification, effective October 2025, p. 1; NCEES Examinee Guide, May 2026, pp. 3, 11, 14, and 16.
What to study first: the four areas by weight
The exam has four knowledge areas. Equipment and systems together account for 44 to 66 of the 80 questions, so that's where most of your hours belong.
| Knowledge area | Questions | Share of range midpoints | What NCEES lists | Practice here |
|---|---|---|---|---|
| HVAC Equipment and Components | 24–36 | about 34% | Cooling towers and fluid coolers; boilers and furnaces; heat exchangers; condensers/evaporators (chillers, VRF, heat pumps, thermal storage); pumps, compressors, and fans (laws, efficiency, selection, cavitation, curves, NPSH); cooling and heating coils; energy recovery | Problems 21–31 |
| HVAC and Refrigeration Distribution and Systems | 20–30 | about 29% | Air distribution and ductwork; air quality and ventilation (filtration, dilution, mixing); fluid distribution and piping (hydronic, fuel oil, fuel gas, steam/condensate); control concepts (valves, dampers, sensors, actuators, temperature reset, PID); refrigeration and refrigeration systems (food processing and storage, refrigerants) | Problems 10–20 |
| HVAC Loads and Psychrometrics | 18–27 | about 26% | Heating and cooling loads; heating and cooling processes; humidification and dehumidification processes, each at sea level, 5,000-ft elevation, and low temperature | Problems 1–9 |
| Supportive Knowledge (HVAC and Refrigeration) | 8–12 | about 11% | Codes and standards; acoustics and vibration control; economic analysis; electrical concepts (power consumption, motor ratings, heat output, amperage) | Problems 32–36 |
The question ranges and topic lists are NCEES's. The "share" column is ours: each range's midpoint divided by the sum of the midpoints (87.5). It's a rough guide to where questions are likely, not an official weight. Because the ranges overlap, no single area is guaranteed to be the largest on your form. NCEES also notes that its listed examples aren't exhaustive.
Source: NCEES HVAC and Refrigeration specification, effective October 2025, pp. 1–2.
Two things in the specification deserve early attention:
- Altitude and low temperature are named outright. Psychrometric processes at 5,000 ft and at low temperature are listed beside sea level. Problems 2 and 5 show why the familiar 1.08 shortcut stops working.
- Most points come from connected methods, not single formulas. Across the whole exam, the same few habits decide the answer: put moist air on a dry-air mass basis, draw the boundary the question asks about (shaft, electrical, or fluid power; compressor or total COP), and carry units through every line.
Studying from an older book? Where each topic moved
Before October 2025, the HVAC exam used a different outline: Principles (28–43 questions) and Applications (42–64 questions). Many review books, courses, and question banks still use that structure. The engineering hasn't changed much, but the grouping and weighting have.
| April 2020 specification (topics since April 2017) | October 2025 specification |
|---|---|
| Basic Engineering Practice: units, economic analysis, electrical concepts | Units run through every area. Economics is now 4C; electrical concepts are 4D. |
| Thermodynamics, Heat Transfer, Fluid Mechanics, Energy/Mass Balances (17–26 questions combined) | No standalone areas. These principles are tested inside load, system, and equipment problems. |
| Psychrometrics (sea level, 5,000-ft elevation) | Area 1B and 1C, which now also name low temperature |
| Heating/Cooling Loads | 1A |
| Equipment: towers, boilers, heat exchangers, condensers/evaporators, pumps/compressors/fans, coils | 3A–3F, which now name thermal storage and cavitation, curves, and NPSH as examples |
| Control system components; basic control concepts | Combined in 2D, which now names sensors, actuators, and PID |
| Refrigerants; refrigeration components; refrigeration systems | Combined in 2E |
| Energy recovery (under Systems) | Moved to Equipment, 3G |
| Air quality and ventilation (under Supportive Knowledge) | Moved to 2B, which now names mixing |
| Fluid distribution, with compressed air among the examples | 2C lists hydronic, fuel oil, fuel gas, and steam/condensate |
| Supportive Knowledge: 3–5 questions | Area 4: 8–12 questions |
Our takeaway: an older book is still fine for learning the methods. Use the October 2025 specification to decide what to study and how much time to give it. If your book says nothing about altitude, low-temperature processes, thermal storage, or NPSH, add those topics yourself.
Sources: April 2020 specification, pp. 1–2; October 2025 specification, pp. 1–2.
36 original practice problems with worked solutions
These are original Castleport problems written to the October 2025 HVAC and Refrigeration topic list. They aren't NCEES questions, they aren't a full-length exam, and they don't cover every listed subtopic in depth. Every problem gives you the data it needs, so you can solve it without the handbook.
Most are single-answer multiple choice. Five ask you to enter a number, two are select-all-that-apply, and one is a matching item. NCEES also uses point-and-click items, which this page doesn't reproduce. On the real exam, every question is scored right or wrong, with no partial credit (Examinee Guide, p. 11).
Work each problem before you open the solution. When you miss one, write down the first step that went wrong. The error log shows what to do with it.
HVAC Loads and Psychrometrics (Problems 1–9)
What this area tests most often:
- Sensible, latent, and total heat are different equations. Sensible uses a temperature change, latent uses a humidity-ratio change, and total uses an enthalpy change. The shortcuts 1.08, 4,840, and 4.5 all assume standard air (0.075 lb/ft³).
- Humidity ratio and enthalpy are per pound of dry air. Mass-weight them, never volume-weight them.
- Altitude and cold air change the mass in a cubic foot. Ask where the airflow was measured.
Problem 1 is at the top of this page.
Problem 2 · The same load at 5,000 ft
HLP-02 · HVAC Loads and Psychrometrics, 1B · Select one
The space from Problem 1 is now at 5,000 ft elevation, where standard barometric pressure is 12.23 psia (sea level: 14.696 psia). It still gets 2,000 cfm at 55°F and is held at 75°F. Take sea-level standard air as 0.075 lb/ft³, assume density scales with pressure at the same temperature, and use cp = 0.24 Btu/(lb·°F). What sensible load does the supply air remove, closest to?
Show answer and solution
Answer and solution
Answer: B. 36,000 Btu/h.
Density at 5,000 ft = 0.075 × (12.23 ÷ 14.696) = 0.0624 lb/ft³. The sensible factor becomes 60 × 0.0624 × 0.24 = 0.899, so Qs = 0.899 × 2,000 × 20 ≈ 35,950 Btu/h, or about 36,000 Btu/h. That's roughly 17% less heat than at sea level for the same cfm. A uses the sea-level 1.08 factor. C divides by the pressure ratio instead of multiplying, as if thinner air carried more heat. D applies the pressure ratio twice.
Try a change: Need to remove the full 43,200 Btu/h at 5,000 ft? Airflow must rise to about 43,200 ÷ (0.899 × 20) ≈ 2,400 cfm.
Source: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, §2 U.S. Standard Atmosphere, Table 1; NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1B.
Problem 3 · Conduction through a layered wall
HLP-03 · HVAC Loads and Psychrometrics, 1A · Select one
A 400 ft² wall has these thermal resistances in series, in h·ft²·°F/Btu: inside air film 0.68, gypsum board 0.45, insulation 13.0, sheathing 0.50, outside air film 0.17. Indoor air is 70°F and outdoor air is 10°F. What is the steady conduction heat loss, closest to?
Show answer and solution
Answer and solution
Answer: A. 1,620 Btu/h.
Resistances in series add: R = 0.68 + 0.45 + 13.0 + 0.50 + 0.17 = 14.8. So U = 1/14.8 = 0.0676 Btu/(h·ft²·°F), and Q = U × A × ΔT = 400 × 60 ÷ 14.8 ≈ 1,620 Btu/h. B counts only the insulation (400 × 60 ÷ 13). The films and layers still resist heat flow. C uses 70°F as the temperature difference instead of 70 − 10 = 60°F. D multiplies by R instead of dividing.
Source: DOE Fundamentals Handbook, Heat Transfer (DOE-HDBK-1012/2-92), Module HT-02, Eqs. 2-2 and 2-3, pp. 3–4. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1A.
Problem 4 · Latent load from infiltration
HLP-04 · HVAC Loads and Psychrometrics, 1C · Select one
Infiltration brings 300 cfm of outdoor air (humidity ratio 0.0157 lb/lb) into a space held at 0.0086 lb/lb. Use standard air. What latent load does the infiltration add, closest to?
Show answer and solution
Answer and solution
Answer: B. 10,300 Btu/h.
Ql = 4,840 × cfm × ΔW = 4,840 × 300 × (0.0157 − 0.0086) = 4,840 × 300 × 0.0071 ≈ 10,300 Btu/h. The 4,840 comes from 4.5 lb/h per cfm times roughly 1,076 Btu/lb of moisture. A treats ΔW × 1,000 as grains and uses the 0.68 grains factor. One lb/lb is 7,000 grains/lb, not 1,000. C doubles the answer. D uses the outdoor humidity ratio alone instead of the difference.
Source: NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1C.
Problem 5 · Low-temperature air carries more mass
HLP-05 · HVAC Loads and Psychrometrics, 1B · Select one
A preheat coil warms 1,500 cfm of outdoor air from 0°F to 70°F. The airflow is measured at the 0°F entering condition, at sea level (14.696 psia). Treat the air as dry with R = 53.35 ft·lbf/(lb·°R) and cp = 0.24 Btu/(lb·°F). What heating rate is required, closest to?
Show answer and solution
Answer and solution
Answer: B. 130,500 Btu/h.
Find the real density at 0°F: ρ = P ÷ (R × T) = (14.696 × 144) ÷ (53.35 × 459.67) ≈ 0.0863 lb/ft³. Then Q = 60 × 0.0863 × 0.24 × 1,500 × 70 ≈ 130,500 Btu/h. Cold air is denser than the 0.075 lb/ft³ behind the 1.08 shortcut, so each cfm carries more mass. A uses the 1.08 factor, which assumes 0.075 lb/ft³ air at about 70°F. C corrects for density in the wrong direction. D leaves out the 60 min/h conversion.
Try a change: If the same 1,500 cfm were measured after the coil, at 70°F, the 1.08 shortcut would be close. Always ask where the airflow was measured.
Source: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, §6 Perfect Gas Relationships; NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1B.
Problem 6 · Total coil load from enthalpy
HLP-06 · HVAC Loads and Psychrometrics, 1B · Select one
A cooling coil handles 4,000 cfm of standard air. Entering enthalpy is 31.5 Btu/lb dry air and leaving enthalpy is 21.5 Btu/lb dry air. What is the total coil load?
Show answer and solution
Answer and solution
Answer: B. 180,000 Btu/h.
Qt = 4.5 × cfm × Δh = 4.5 × 4,000 × (31.5 − 21.5) = 180,000 Btu/h, or 15 tons. The 4.5 is 60 min/h × 0.075 lb/ft³, the mass flow in lb/h per cfm. A uses the 1.08 sensible factor with Δh as if it were a temperature change. C leaves out the 4.5 factor. D drops a zero.
Source: NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1B.
Problem 7 · Sensible heat ratio
HLP-07 · HVAC Loads and Psychrometrics, 1A · Select one
The coil in Problem 6 cools the air from 78°F to 55°F while removing 180,000 Btu/h total. Using the 1.08 factor for the sensible part, what is the sensible heat ratio (SHR), closest to?
Show answer and solution
Answer and solution
Answer: A. 0.55.
Sensible load = 1.08 × 4,000 × (78 − 55) = 99,360 Btu/h. SHR = sensible ÷ total = 99,360 ÷ 180,000 ≈ 0.55. The other 45% is latent load: moisture condensed on the coil. B is the latent fraction, 1 − SHR. C divides total by sensible. An SHR can't exceed 1. D divides the temperature drop by the entering temperature, which isn't an energy ratio.
Source: ASHRAE Terminology, sensible heat ratio; NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1A.
Problem 8 · Steam needed for humidification
HLP-08 · HVAC Loads and Psychrometrics, 1C · Select one
A steam humidifier raises 3,000 cfm of standard air from 15 to 40 grains of moisture per pound of dry air. There are 7,000 grains in a pound. How much water must the humidifier add, closest to?
Show answer and solution
Answer and solution
Answer: A. 48 lb/h.
Dry-air mass flow is 4.5 × 3,000 = 13,500 lb/h. Water added = 13,500 × (40 − 15) ÷ 7,000 ≈ 48 lb/h. B forgets to convert cfm to lb/h (leaves out the 4.5). C uses the final 40 grains instead of the 25-grain increase. D never converts grains to pounds.
Source: NEBB Fundamental Formula Chart (2025), Heat Transfer (Air) and Engineering Constants; ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, §5 Humidity Parameters, Eq. (7). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1C.
Problem 9 · What a wet cooling coil does
HLP-09 · HVAC Loads and Psychrometrics, 1C · Select all that apply
Air passes over a cooling coil whose surface is colder than the entering air's dew point. Compared with the entering air, which statements about the leaving air are true? Select all that apply.
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Answer and solution
Answer: A, B and C.
Below the dew point, water vapor condenses on the coil. The air leaves cooler (A), with less moisture (B), and with less total energy (C). This is cooling with dehumidification. D describes sensible-only cooling, which happens when the coil surface stays above the dew point. E is backwards. Removing moisture lowers the dew point.
Scored all-or-nothing, like NCEES multiple-correct items: you need A, B, and C and nothing else.
Source: EPA 402-F-13053, Moisture Control Guidance, pp. 15–19; ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, §5 Humidity Parameters, Eq. (7). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 1C.
HVAC and Refrigeration Distribution and Systems (Problems 10–20)
What this area tests most often:
- Ducts and ventilation: velocity from cfm and area, velocity pressure, outdoor-air fraction, and dilution airflow.
- Piping: water (500 × gpm × ΔT), steam (latent heat per pound), and fuel gas (input ÷ heating value).
- Controls: a sensor measures, the controller compares the reading with the setpoint, and its output moves an actuator on a valve or damper. Reset changes the setpoint as conditions change. A warmer chilled-water or supply-air setpoint can save chiller energy, but it can also cost humidity control and fan or pump energy, so it isn't automatically a saving.
- Refrigeration cycles: identify which component each enthalpy difference crosses, and remember the expansion valve doesn't change enthalpy.
Problem 10 · Duct velocity pressure
DS-01 · HVAC and Refrigeration Distribution and Systems, 2A · Select one
A 16 in. × 12 in. rectangular duct carries 2,000 cfm of standard air. What is the velocity pressure, closest to?
Show answer and solution
Answer and solution
Answer: A. 0.14 in. w.g.
Area = (16 × 12) ÷ 144 = 1.333 ft², so V = 2,000 ÷ 1.333 = 1,500 fpm. For standard air, V = 4,005√VP, so VP = (1,500 ÷ 4,005)² ≈ 0.14 in. w.g. B forgets to square the ratio. C doubles the squared result. D inverts the ratio (4,005 ÷ 1,500)².
Try a change: If you'd used the area in square inches, you'd get a velocity of about 10 fpm. Convert to ft² first.
Source: NEBB Fundamental Formula Chart (2025), Airflow & Velocity. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2A.
Problem 11 · Mix on a dry-air mass basis
DS-02 · HVAC and Refrigeration Distribution and Systems, 2B · Enter a number
Two moist-air streams mix steadily at sea level with no heat transfer, condensation, or added moisture. Stream A: 780 cfm, specific volume 13.0 ft³/lb dry air, humidity ratio 0.00400 lb/lb. Stream B: 1,800 cfm, specific volume 15.0 ft³/lb dry air, humidity ratio 0.01200 lb/lb. Enter the mixed humidity ratio in lb water/lb dry air, to five decimal places.
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Answer and solution
Answer: 0.00933 lb water/lb dry air.
Convert each stream to dry-air mass with its own specific volume: 780 ÷ 13.0 = 60 lb/min and 1,800 ÷ 15.0 = 120 lb/min. Then mass-weight the humidity ratios: W = (60 × 0.00400 + 120 × 0.01200) ÷ 180 = 0.00933 lb/lb. An answer of 0.00958 weights by the raw cfm values. Volume isn't mass when the streams have different densities. An answer of 0.00800 is a simple average that ignores flow. An answer of 0.01600 adds the two ratios.
Source: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, §5 Humidity Parameters, Eq. (7); Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2B.
Problem 12 · Outdoor-air fraction from temperatures
DS-03 · HVAC and Refrigeration Distribution and Systems, 2B · Select one
An air handler's sensors read: mixed air 62°F, return air 75°F, outdoor air 45°F. Assuming well-mixed air and accurate sensors, what fraction of the supply is outdoor air, closest to?
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Answer and solution
Answer: A. 43%.
%OA = (T_return − T_mixed) ÷ (T_return − T_outdoor) = (75 − 62) ÷ (75 − 45) = 13 ÷ 30 ≈ 43%. B is the return-air fraction, 1 − 0.43. C divides 13 by the return temperature. D divides 13 by the mixed-air temperature.
Try a change: This method breaks down when outdoor and return temperatures are close, because the denominator gets tiny and sensor error dominates.
Source: NEBB Fundamental Formula Chart (2025), Air Temperature equations (percent outdoor air, mixed-air temperature). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2B.
Problem 13 · Dilution ventilation
DS-04 · HVAC and Refrigeration Distribution and Systems, 2B · Select one
A process releases 1.2 ft³/h of a gaseous contaminant into a room. The concentration must stay at or below 50 ppm (by volume), and the outdoor air brought in contains none. Assume perfect mixing (a mixing factor of 1) and steady state. What outdoor airflow is required?
Show answer and solution
Answer and solution
Answer: A. 400 cfm.
Q = G ÷ C = 1.2 ft³/h ÷ (50 × 10⁻⁶) = 24,000 ft³/h. Divide by 60: 400 cfm. B leaves the answer in ft³/h and calls it cfm. C treats ppm as parts per thousand. D multiplies by a mixing factor of 5. Real designs do apply a factor for imperfect mixing, but this problem says to assume 1.
Source: OSHA Technical Manual, Sec. III Ch. 3, Appendix III:3-1, general exhaust (dilution) ventilation. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2B.
Problem 14 · Hydronic flow rate
DS-05 · HVAC and Refrigeration Distribution and Systems, 2C · Enter a number
A hot-water heating loop delivers 240,000 Btu/h with a 20°F temperature drop. Enter the required water flow in gpm.
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Answer and solution
Answer: 24 gpm.
Q = 500 × gpm × ΔT, so gpm = 240,000 ÷ (500 × 20) = 24 gpm. The 500 is about 8.33 lb/gal × 60 min/h × 1 Btu/(lb·°F), so it applies to water, not glycol. An answer of 480 multiplies by ΔT instead of dividing. An answer of 2.4 is a decimal slip.
Source: NEBB Fundamental Formula Chart (2025), Heat Transfer (Hydronic). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2C.
Problem 15 · Steam condensate rate
DS-06 · HVAC and Refrigeration Distribution and Systems, 2C · Select one
A unit heater delivers 500,000 Btu/h using saturated steam at 15 psig. The condensate leaves as saturated liquid. Take the latent heat of vaporization (hfg) at this pressure as 945 Btu/lb. How much steam does the heater condense, closest to?
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Answer and solution
Answer: A. 529 lb/h.
When steam condenses to saturated liquid, each pound gives up its latent heat, hfg. Steam flow = 500,000 ÷ 945 ≈ 529 lb/h. B divides by hg (about 1,160 Btu/lb), the total enthalpy of the vapor, rather than the heat actually released. C uses a 1,000 Btu/lb rule of thumb instead of the stated value. D divides by hf (about 218 Btu/lb), the liquid's enthalpy.
Source: DOE, Improving Steam System Performance, 2nd ed., pp. 2–3. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2C.
Problem 16 · Fuel gas flow to a furnace
DS-07 · HVAC and Refrigeration Distribution and Systems, 2C · Select one
A furnace delivers 80,000 Btu/h of heat at 80% efficiency (efficiency = output ÷ input). The natural gas has a heating value of 1,030 Btu/ft³. What gas flow does the furnace burn, closest to?
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Answer and solution
Answer: A. 97 cfh.
Input = output ÷ efficiency = 80,000 ÷ 0.80 = 100,000 Btu/h. Gas flow = 100,000 ÷ 1,030 ≈ 97 ft³/h. B uses the output, not the input. C multiplies by the efficiency. D divides 100,000 by 800 instead of 1,030.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2C.
Problem 17 · Getting rid of controller offset
DS-08 · HVAC and Refrigeration Distribution and Systems, 2D · Select one
A proportional-only controller holds discharge-air temperature. Under a steady load, it settles 2°F above setpoint and stays there. Which change is the standard way to eliminate this steady error?
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Answer and solution
Answer: A. Add integral action (make it a PI controller).
A proportional-only controller needs some error to produce any output, so it settles with an offset. Integral action keeps adding output as long as error persists, which drives the steady error to zero. B shrinks the offset but doesn't remove it, and too much gain can make the loop oscillate. C responds to how fast the error changes. A steady error doesn't change. D slows the response without addressing the offset.
Source: Seborg et al., Process Dynamics and Control, Ch. 8 slides 7, 8, and 26 (UC Santa Barbara). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2D.
Problem 18 · Outdoor-air reset schedule
DS-09 · HVAC and Refrigeration Distribution and Systems, 2D · Enter a number
A hot-water reset schedule is linear: 180°F supply at 0°F outdoor air and 120°F supply at 60°F outdoor air. Enter the hot-water setpoint, in °F, when it is 30°F outside.
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Answer and solution
Answer: 150 °F.
The setpoint drops 60°F over a 60°F rise in outdoor temperature, or 1°F per °F. At 30°F outdoors: 180 − 30 × 1 = 150°F. An answer of 210 adds instead of subtracting. Supply water gets cooler as it gets warmer outside.
Source: PNNL-21569, Energy Savings Modeling of Standard Commercial Building Re-tuning Measures (2012), measure W04, hot-water supply temperature reset. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2D.
Problem 19 · Refrigerant flow for a 10-ton system
DS-10 · HVAC and Refrigeration Distribution and Systems, 2E · Select one
A vapor-compression system provides 10 tons of refrigeration (1 ton = 12,000 Btu/h). Given refrigerant enthalpies: evaporator outlet h1 = 104 Btu/lb, compressor outlet h2 = 118 Btu/lb, condenser outlet h3 = 40 Btu/lb. The expansion valve is adiabatic. What refrigerant mass flow is required?
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Answer and solution
Answer: A. 1,875 lb/h.
An adiabatic expansion valve does no work, so h4 = h3 = 40 Btu/lb. Refrigerating effect = h1 − h4 = 104 − 40 = 64 Btu/lb. Mass flow = 120,000 ÷ 64 = 1,875 lb/h. As a check, compressor work is 1,875 × (118 − 104) = 26,250 Btu/h, so COP based on compressor work is 64 ÷ 14 ≈ 4.57. B divides by h1 instead of the enthalpy change across the evaporator. C uses h2 − h4, which is the condenser's heat rejection per pound, not the cooling. D divides by the compressor work per pound.
Try a change: A COP based on total electrical input, including fans, pumps, and motor losses, will be lower than 4.57. Read which boundary the question asks for.
Source: ASHRAE Handbook—Fundamentals (2025), Ch. 2, §1.5 Eq. (15)–(16) and §2.2 Eqs. (42a)–(42d); FEMP, How to Buy an Energy-Efficient Air-Cooled Electric Chiller (2003), Metric Conversion note. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2E.
Problem 20 · Product freezing load
DS-11 · HVAC and Refrigeration Distribution and Systems, 2E · Select one
A freezer cools 2,000 lb of product from 40°F to 0°F. The product freezes at 28°F. Use specific heat 0.80 Btu/(lb·°F) above freezing, 0.42 Btu/(lb·°F) below freezing, and a latent heat of fusion of 100 Btu/lb. What total heat must be removed, closest to?
Show answer and solution
Answer and solution
Answer: A. 242,700 Btu.
Split it into three steps. Above freezing: 2,000 × 0.80 × (40 − 28) = 19,200 Btu. Freezing: 2,000 × 100 = 200,000 Btu. Below freezing: 2,000 × 0.42 × (28 − 0) = 23,520 Btu. Total ≈ 242,700 Btu. B leaves out the latent heat, which is most of the load. C uses the above-freezing specific heat across the whole 40°F drop. D counts only the latent heat.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 2E.
HVAC Equipment and Components (Problems 21–31)
What this area tests most often:
- Match capacity to both loads. A coil's total rating means little until you split it into sensible and latent capacity.
- Reconcile both sides of a heat exchanger before you calculate an area or an LMTD.
- Separate fluid, shaft, and electrical power. Divide by pump or fan efficiency to reach the shaft, and by motor efficiency to reach the electrical input.
- Fan and pump laws: flow follows speed, pressure follows speed squared, and power follows speed cubed, under the same-equipment, same-system assumptions.
- Cooling towers: range is the water's temperature drop. Approach is leaving water temperature minus entering wet-bulb.
Problem 21 · Cooling tower range, approach, and heat rejection
EQ-01 · HVAC Equipment and Components, 3A · Select one
A cooling tower cools 600 gpm of water from 95°F to 85°F. The entering air wet-bulb temperature is 78°F. What heat does the tower reject, and what is its approach?
Show answer and solution
Answer and solution
Answer: A. 3.0 million Btu/h; 7°F.
Range is the water's temperature drop: 95 − 85 = 10°F. Heat rejected = 500 × 600 × 10 = 3,000,000 Btu/h. Approach is leaving water temperature minus entering wet-bulb: 85 − 78 = 7°F. B measures approach from the entering water temperature. C uses 95 − 78 as the range. D swaps range and approach.
Source: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 40 Cooling Towers, §1 Principle of Operation and §2 Design Conditions; NEBB Fundamental Formula Chart (2025), Heat Transfer (Hydronic). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3A.
Problem 22 · Boiler fuel input
EQ-02 · HVAC Equipment and Components, 3B · Select one
A boiler must deliver 2,000,000 Btu/h of useful heat at 82% efficiency (efficiency = output ÷ input). What fuel input is required, closest to?
Show answer and solution
Answer and solution
Answer: A. 2.44 million Btu/h.
Input = output ÷ efficiency = 2,000,000 ÷ 0.82 ≈ 2,440,000 Btu/h. Input has to be larger than output. B multiplies by the efficiency. C ignores the efficiency. D is a decimal slip.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3B.
Problem 23 · Log mean temperature difference
EQ-03 · HVAC Equipment and Components, 3C · Select one
In a counterflow heat exchanger, hot water cools from 180°F to 140°F and cold water warms from 100°F to 130°F. What is the log mean temperature difference (LMTD), closest to?
Show answer and solution
Answer and solution
Answer: A. 44.8°F.
In counterflow, pair the hot inlet with the cold outlet and the hot outlet with the cold inlet. ΔT1 = 180 − 130 = 50°F and ΔT2 = 140 − 100 = 40°F. LMTD = (50 − 40) ÷ ln(50/40) ≈ 44.8°F. B is the arithmetic mean. It's close here because the two end differences are similar, but the gap grows fast when they aren't. C pairs the ends as if the flow were parallel (80°F and 10°F). D uses only one end.
Source: DOE Fundamentals Handbook, Heat Transfer (DOE-HDBK-1012/2-92), Module HT-02, Eqs. 2-2 and 2-3, pp. 3–4. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3C.
Problem 24 · Balance both sides of a heat exchanger
EQ-04 · HVAC Equipment and Components, 3C · Select one
An insulated water-to-water heat exchanger runs steadily. Hot water, 7,200 lb/h, enters at 160°F and leaves at 120°F. Cold water, 10,800 lb/h, enters at 60°F. Use cp = 1.00 Btu/(lb·°F) for both streams. What is the cold-water outlet temperature, closest to?
Show answer and solution
Answer and solution
Answer: B. 86.7°F.
The hot side gives up 7,200 × 1.00 × (160 − 120) = 288,000 Btu/h. The cold side picks up the same amount: T_out = 60 + 288,000 ÷ (10,800 × 1.00) ≈ 86.7°F. A subtracts the heat from the cold stream. C assumes equal flows, so the cold side rises 40°F. D swaps the two mass flows.
Source: DOE Fundamentals Handbook, Heat Transfer (DOE-HDBK-1012/2-92), Module HT-02, Eqs. 2-2 and 2-3, pp. 3–4; Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3C.
Problem 25 · Chiller kW/ton to COP
EQ-05 · HVAC Equipment and Components, 3D · Enter a number
A chiller is rated at 0.60 kW/ton. One ton of refrigeration is 12,000 Btu/h, or 3.517 kW. Enter its COP, to two decimal places.
Show answer and solution
Answer and solution
Answer: 5.86.
COP is cooling output divided by power input in the same units. One ton is 3.517 kW of cooling, so COP = 3.517 ÷ 0.60 ≈ 5.86. An answer of 20 is the EER: 12,000 Btu/h ÷ 600 W = 20 Btu/(W·h). EER mixes units; COP doesn't. An answer of 0.17 inverts the ratio.
Source: FEMP, How to Buy an Energy-Efficient Air-Cooled Electric Chiller (2003), Metric Conversion note. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3D.
Problem 26 · Ice storage capacity
EQ-06 · HVAC Equipment and Components, 3D · Select one
An ice thermal storage tank melts 10,000 lb of ice during the day. Use a latent heat of fusion of 144 Btu/lb and 1 ton-hour = 12,000 Btu. How much cooling does the melting provide?
Show answer and solution
Answer and solution
Answer: A. 120 ton-hours.
10,000 × 144 = 1,440,000 Btu. Divide by 12,000 Btu per ton-hour: 120 ton-hours. B never converts Btu to ton-hours. C divides by 144,000. D treats a ton-hour as 1,000 Btu.
Source: FEMP, How to Buy an Energy-Efficient Air-Cooled Electric Chiller (2003), Metric Conversion note; Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3D.
Problem 27 · Fan laws after a speed change
EQ-07 · HVAC Equipment and Components, 3E · Select all that apply
A fan needs 5.00 hp of shaft power at 900 rpm. Its speed is raised to 990 rpm. Assume the same fan, same air density, a fixed system curve, and unchanged fan efficiency. Motor and drive efficiencies are not given. Which statements are true? Select all that apply.
Show answer and solution
Answer and solution
Answer: A, B and C.
Speed ratio = 990 ÷ 900 = 1.10. Airflow scales with speed (1.10), pressure with speed squared (1.21), and power with speed cubed: 5.00 × 1.10³ ≈ 6.66 hp. A 10% speed increase costs about 33% more power. D applies the airflow exponent to power. E crosses into the motor and drive, whose efficiencies aren't given. The fan law gives shaft power only.
Scored all-or-nothing: A, B, and C, and nothing else.
Source: DOE, Improving Fan System Performance, p. 4, Rotational Speed. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3E.
Problem 28 · Net positive suction head available
EQ-08 · HVAC Equipment and Components, 3E · Select one
A pump draws 60°F water from an open tank at sea level. Use atmospheric pressure head = 33.9 ft and vapor pressure head = 0.59 ft. The water surface is 5 ft above the pump centerline, and suction-line friction is 3 ft. The pump requires 8 ft NPSH. What is NPSH available, and is there margin?
Show answer and solution
Answer and solution
Answer: A. 35.3 ft; yes, ample margin.
NPSHA = atmospheric head + static head − friction − vapor pressure head = 33.9 + 5 − 3 − 0.59 ≈ 35.3 ft. That's far above the 8 ft required, so cavitation isn't expected. B subtracts the 5 ft as if it were a suction lift. Here the water is above the pump, so it helps. C adds friction instead of subtracting it. D leaves out atmospheric pressure, the biggest term.
Source: DOE Fundamentals Handbook, Fluid Flow (DOE-HDBK-1012/3-92), Module 3, pp. 48–49, Eq. 3-19; DOE, Improving Pumping System Performance, 2nd ed., pp. 39–40 (NPSHA and NPSHR). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3E.
Problem 29 · Stop at the shaft-power boundary
EQ-09 · HVAC Equipment and Components, 3E · Select one
A pump moves 100 gpm of water (specific gravity 1.00) against 80 ft of total head. Pump efficiency is 70% and motor efficiency is 90%. What shaft power must be delivered to the pump, closest to?
Show answer and solution
Answer and solution
Answer: C. 2.89 hp.
Fluid power = gpm × head × SG ÷ 3,960 = 100 × 80 ÷ 3,960 ≈ 2.02 hp. Shaft power = fluid power ÷ pump efficiency = 2.02 ÷ 0.70 ≈ 2.89 hp. A multiplies by pump efficiency. B stops at fluid power. D also divides by motor efficiency, which answers an electrical-input question.
Source: DOE, Improving Pumping System Performance, 2nd ed., p. 9 (fluid power and pump efficiency). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3E.
Problem 30 · Pick a coil that meets both loads
EQ-10 · HVAC Equipment and Components, 3F · Select one
A space needs 24,000 Btu/h of sensible cooling and 6,000 Btu/h of latent cooling. Four coils are rated at the same design conditions. Which one meets both loads without extra dehumidification? A: 32,000 Btu/h total, SHR 0.80. B: 35,000 Btu/h total, SHR 0.90. C: 29,000 Btu/h total, SHR 0.78. D: 30,000 Btu/h total, SHR 0.90.
Show answer and solution
Answer and solution
Answer: A. Coil A.
Sensible capacity = SHR × total; latent is the rest. A: 25,600 sensible and 6,400 latent, so it meets both. B: 31,500 and 3,500 (short on latent). C: 22,620 and 6,380 (short on sensible). D: 27,000 and 3,000 (short on latent). B has the biggest total, but its high SHR leaves only 3,500 Btu/h for moisture. C handles the latent load but falls short on sensible. D falls short on latent capacity.
Source: ASHRAE Terminology, sensible heat ratio. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3F.
Problem 31 · Energy recovery wheel leaving temperature
EQ-11 · HVAC Equipment and Components, 3G · Select one
A sensible energy recovery wheel has an effectiveness of 0.70, defined here as (supply air leaving the wheel − outdoor air) ÷ (return air − outdoor air), with equal airflows. Outdoor air is 20°F and return air is 70°F. What is the supply air temperature leaving the wheel?
Show answer and solution
Answer and solution
Answer: A. 55°F.
T = 20 + 0.70 × (70 − 20) = 20 + 35 = 55°F. The wheel recovers 70% of the possible temperature rise. B multiplies the return temperature by 0.70. C finds the 35°F rise but forgets to add it to the outdoor temperature. D assumes the wheel recovers all the available heat (effectiveness 1.0). Real wheels don't.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 3G.
Supportive Knowledge (Problems 32–36)
What this area tests most often:
- Codes and standards: no standards are supplied on this exam, so know what each major document governs rather than its tables.
- Sound: decibels add logarithmically. Two equal sources add about 3 dB.
- Vibration: isolators help only when the forcing frequency is more than √2 times their natural frequency.
- Economics: a positive present worth at the stated rate beats a payback rule of thumb.
- Electrical: motor input = output ÷ efficiency, and where the motor sits decides how much of that becomes heat in the air.
Problem 32 · Which ASHRAE standard covers what?
SK-01 · Supportive Knowledge (HVAC and Refrigeration), 4A · Match each pair
Match each ASHRAE standard to its subject. Standards: 15, 34, 55, 62.1, 90.1. Subjects: (a) building energy efficiency; (b) designation and safety classification of refrigerants; (c) ventilation and acceptable indoor air quality; (d) safety of refrigeration systems; (e) thermal environmental conditions for human occupancy.
| Standard | Subject |
|---|---|
| 15 | |
| 34 | |
| 55 | |
| 62.1 | |
| 90.1 |
Show answer and solution
Answer and solution
Answer: 15 → (d) safety of refrigeration systems; 34 → (b) designation and safety classification of refrigerants; 55 → (e) thermal environmental conditions for human occupancy; 62.1 → (c) ventilation and acceptable indoor air quality; 90.1 → (a) building energy efficiency.
15 is the refrigeration safety standard and 34 assigns refrigerant numbers and safety groups. 55 covers thermal comfort, 62.1 covers ventilation, and 90.1 is the commercial energy standard. None of these documents is supplied on the PE Mechanical exam, so know what each one governs, not its tables.
Scored all-or-nothing: all five pairs must be correct.
Source: ASHRAE, read-only versions of ASHRAE standards (current titles); ASHRAE, Titles, Purposes, and Scopes (Standard 34). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 4A.
Problem 33 · Adding two equal sound levels
SK-02 · Supportive Knowledge (HVAC and Refrigeration), 4B · Enter a number
Two identical fans each produce 80 dB at a listener's position. Enter the combined sound level, in dB, when both run.
Show answer and solution
Answer and solution
Answer: 83 dB.
Decibels add logarithmically: L = 10 log₁₀(10⁸ + 10⁸) ≈ 83 dB. Two equal sources double the sound power, which adds about 3 dB. An answer of 160 adds the decibels arithmetically.
Source: ASHRAE Handbook—Fundamentals (2025), Ch. 8 Sound and Vibration, Combining Sound Levels, Eq. (10) and Table 3; OSHA Technical Manual, Sec. III Ch. 5 (Noise), II.B.5 and II.B.7. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 4B.
Problem 34 · Vibration isolation efficiency
SK-03 · Supportive Knowledge (HVAC and Refrigeration), 4B · Select one
A fan turns at 1,200 rpm (20 Hz) on spring isolators with a natural frequency of 5 Hz. Damping is light; use the undamped model, transmissibility T = 1 ÷ |(f/fn)² − 1|. What percentage of the vibration force is isolated, closest to?
Show answer and solution
Answer and solution
Answer: A. 93%.
Frequency ratio r = 20 ÷ 5 = 4. T = 1 ÷ (16 − 1) ≈ 0.067, so about 6.7% of the force gets through and about 93% is isolated. Isolation only happens when r is above √2 ≈ 1.41. B uses 1 − 1/r, which isn't the transmissibility relation. C reports the transmitted fraction as the isolated fraction. D would be true near r = 1 (resonance), not at r = 4.
Source: University of Iowa 53/58:153, Lecture 9, p. 10 (vibration isolation). Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 4B.
Problem 35 · Is the retrofit worth it?
SK-04 · Supportive Knowledge (HVAC and Refrigeration), 4C · Select one
An HVAC retrofit costs $40,000 now and saves $6,000 per year, received at the end of each year, for 10 years. The discount rate is 6%, and the uniform present value factor (P/A, 6%, 10) is 7.3601. What is the net present value, and should it go ahead on that basis?
Show answer and solution
Answer and solution
Answer: A. About +$4,160; yes.
Present value of savings = $6,000 × 7.3601 ≈ $44,160. NPV = $44,160 − $40,000 ≈ +$4,160. A positive NPV at the stated discount rate means the savings more than cover the cost. B ignores the time value of money ($60,000 − $40,000). C flips the sign. D uses simple payback, which ignores the savings after year 6.7 and the time value of money. A payback under the 10-year life doesn't settle the question either way.
Source: NIST Handbook 135 (2022), §3.2.2.1 Annually Recurring Uniform Amounts, p. 26, and Table 3-2. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 4C.
Problem 36 · Motor input power
SK-05 · Supportive Knowledge (HVAC and Refrigeration), 4D · Select one
A 20 hp motor drives a supply fan at full load. Motor efficiency is 90%. Both the motor and the fan sit in the airstream, so treat all of the motor's electrical input as heat added to the air. What is the electrical input, closest to?
Show answer and solution
Answer and solution
Answer: A. 16.6 kW.
Shaft output = 20 × 0.746 = 14.9 kW. Input = output ÷ efficiency = 14.9 ÷ 0.90 ≈ 16.6 kW. With the motor in the airstream, that full 16.6 kW ends up as heat in the air. B is the shaft output, which ignores motor losses. C multiplies by efficiency. D divides horsepower by efficiency and calls it kW, skipping the 0.746 conversion.
Try a change: If the motor sat outside the airstream, only the 14.9 kW of shaft power would reach the air.
Source: DOE fact sheet, Determining Electric Motor Load and Efficiency, Eq. 2, p. 4. Topic: NCEES PE Mechanical: HVAC and Refrigeration specification, 4D.
How to read your results
Treat your percentage as feedback on these 36 problems, not an NCEES score or a prediction of passing. Five to eleven problems per area can show which methods you can't yet explain, but they can't measure an area precisely.
NCEES counts correct answers, converts the count to a scaled score to adjust for differences between exam forms, and compares it with a standard set by subject-matter experts. It doesn't publish the passing score, so no practice percentage converts into an exam result. (NCEES exam scoring; Examinee Guide, p. 14)
Your 12-week study plan
Assumptions: about 10 focused hours a week (120 hours total), one exam, and a full-time job. This is our planning example, weighted toward the larger NCEES areas. It isn't an NCEES requirement, and no number of hours guarantees a pass.
| Week | Focus | Do this | Finish the week with |
|---|---|---|---|
| 1 | Setup and baseline | Confirm your test date. Download the specification and, from MyNCEES, the PE Mechanical Reference Handbook. Work all 36 problems before opening the solutions. | An error log and a ranked list of the areas where you missed the most |
| 2 | Psychrometrics | Sea level, 5,000 ft, and low-temperature processes; mixing on a dry-air basis; humidification. Rework Problems 1, 2, 5, 6, 8, 9, and 11. | Problems 1 and 2 reworked at an altitude you choose |
| 3 | Loads | Conduction, infiltration, ventilation loads, and sensible heat ratio. Rework Problems 3, 4, and 7. | A one-page load summary with handbook search words for each relation |
| 4 | Air distribution and air quality | Duct velocity and pressure, outdoor-air fraction, dilution, filtration. Rework Problems 10, 12, and 13. | Duct and ventilation lookups logged with their search words |
| 5 | Piping | Hydronic flow, steam and condensate, fuel gas and fuel oil. Rework Problems 14–16. | A steam-table lookup you can finish in under two minutes |
| 6 | Controls and refrigeration systems | Control loops, reset, PID; refrigeration cycles; food storage loads. Rework Problems 17–20. | A refrigeration cycle sketched with all four state points labeled |
| 7 | Pumps, compressors, and fans | Fan and pump laws, efficiency, curves, NPSH, cavitation. Rework Problems 27–29. | Fan-law and NPSH problems you can set up without notes |
| 8 | Towers, boilers, heat exchangers, chillers | Range and approach, efficiency, LMTD and two-stream balances, kW/ton and COP, heat pumps and VRF, thermal storage. Rework Problems 21–26. | A clean units chain: Btu/h, MBH, tons, kW, and ton-hours |
| 9 | Coils, energy recovery, and Supportive Knowledge | Coil selection, recovery effectiveness; standards, acoustics, vibration, economics, electrical. Rework Problems 30–36. | A formula map with the handbook search words that found each relation |
| 10 | Full rehearsal | Sit 80 fresh problems, such as the NCEES practice exam, in 8 hours. Submit the first half before starting the second, and use only the handbook and your approved calculator. | Results logged by area and error type |
| 11 | Fix the leaks | Analyze the rehearsal. Spend most of the week on the areas that cost the most questions, then check each fix on a problem you haven't seen. | A short repair list based on work you actually did |
| 12 | Timed blocks and logistics | Work Block A and Block B below. Confirm your appointment, ID, and calculator. Nothing new the day before. | An exam-day plan and a final list of methods to review |
Week 12 timed blocks: Block A is the odd-numbered problems (1, 3, 5 … 35) and Block B is the even-numbered problems (2, 4, 6 … 36). Each block has 18 problems, so give it 18 × 6 = 108 minutes. These are problems you've already studied, so this is method and pacing practice, not a fresh full-length exam.
Adjusting the plan:
- Fewer hours a week? Stretch the calendar rather than cutting areas. At 6 hours a week, the same plan takes about 20 weeks.
- Only 8 weeks? Combine Weeks 2–3, 4–5, 6–7, and 8–9, keep the full rehearsal and the repair week, and expect about 15 hours a week. If that isn't realistic, consider a later test date.
- Already strong in an area? Keep a short review there and move the saved time to your weakest area. Don't drop any area completely.
- Retaking the exam? Start from your diagnostic report. It shows your performance by knowledge area on a 0–15 scale compared with the average of passing examinees. Weigh both the gap and how many questions the area carries (Examinee Guide, p. 20).
Turn each miss into your next study task
| What went wrong | Next step | How you know it's fixed |
|---|---|---|
| Used volume where mass was needed | Rewrite the balance per pound of dry air or per lb/h, with units on every line | The units cancel to what the question asks for |
| Picked the wrong equation or process | Write one sentence on what the problem asks for and which process applies, then rework a version with one input changed | You can say why the wrong equation doesn't fit |
| Answered the wrong boundary | Sketch the system and circle the boundary: fluid, shaft, or electrical; compressor or total input | You can name what's inside and outside the boundary |
| Slow or wrong handbook lookup | Repeat the lookup and record the section and the search words that worked | The same lookup is faster next time |
| Arithmetic or sign error | Redo it from scratch, then check magnitude and direction | The check agrees without copying the first attempt |
| Ran out of time on a familiar problem | Note where the time went, then practice flagging and returning during a timed block | Your average drops toward 6 minutes |
Error log fields: problem or source · date · first attempt or repeat · minutes spent · your answer · first wrong step · error type (from the table) · corrected method · handbook section and search words · date of your next attempt on a fresh problem.
Log lucky guesses too. And remember that getting a problem right the second time shows you remember it; getting a new problem right shows you can use the method.
If your exam is next week
Don't start a new system. Pick the two errors you repeat most and your two slowest lookups, fix those, and do one timed block. Confirm your logistics. If large gaps remain, you can reschedule your Pearson appointment at least 48 hours ahead for a $50 fee (Examinee Guide, p. 6).
Practice with the only reference you'll have
On exam day, the NCEES PE Mechanical Reference Handbook appears on screen as a searchable PDF. The Examinee Guide's format table lists no design standards for any PE Mechanical exam, so no ASHRAE handbook, SMACNA manual, or code book will be there (Examinee Guide, pp. 10 and 16). You can download the current handbook free from your MyNCEES account.
Two details change how you practice:
- Use the search box, not Ctrl+F. The exam viewer has its own search box on the left side, and Ctrl+F doesn't work. NCEES posts a video on searching the on-screen references.
- Learn locations and assumptions, not page numbers. Find where each relation lives, then read its variable definitions and units before you use it.
A five-minute lookup drill: pick a problem, name the relation you need in plain words, search for that phrase, read the variable definitions and units beside the equation, then apply it. Write down the search words that worked.
Try these lookups in your copy of the handbook and record what you find:
| Find this | Record |
|---|---|
| Moist-air properties and psychrometric chart information, including altitude | The pressure or elevation each chart or table assumes, and its units |
| A sensible heat relation for air | The density and specific heat it assumes |
| Steam and refrigerant property tables | Which column is hf, hfg, and hg |
| Refrigeration cycle and COP relations | The state labels and what the COP is based on |
| Fan and pump laws and power relations | The constant (such as 3,960) and the units it requires |
| Engineering economics factors | Cash-flow timing assumptions and factor notation |
Exam-day rules that change how you practice
- About 6 minutes per question on average. That's 480 minutes ÷ 80 questions, a budgeting guide rather than a per-question limit.
- Two sections, one clock. After roughly half the questions, you review and submit them, and you can't go back. The sections aren't timed separately, so time saved early carries forward.
- Breaks: the optional 50-minute break comes after you submit the first section. Unused break time doesn't add to your exam time, and unscheduled breaks come out of your exam time.
- No penalty for wrong answers, so answer every question. Some questions are unscored pretest items that look like all the others.
- One approved calculator. For 2026 exams, the approved models are the HP 33s and HP 35s, any Casio fx-115 or fx-991 model, and any TI-30X or TI-36X model. An on-screen TI-30XS is also available. Practice on the exact model you'll bring.
- Scratch work: Pearson supplies reusable booklets and markers. The current Examinee Guide does not state a fixed quantity.
- Arrive 30 minutes early, and make sure the first and last names on your appointment confirmation match your ID.
Sources: NCEES Examinee Guide, May 2026, pp. 8–12; NCEES 2026 calculator list, p. 1.
Quick answers
Which PE Mechanical exam should I take?
NCEES offers three separate PE Mechanical exams: HVAC and Refrigeration, Machine Design and Materials, and Thermal and Fluid Systems. Each has 80 questions and is offered year-round. Pick the one that matches the work you do. (NCEES PE Mechanical)
Can I bring ASHRAE or other code books?
No. Only the NCEES handbook is supplied, and personal references aren't allowed in the testing room (Examinee Guide, pp. 8 and 16).
What's the pass rate?
In the NCEES table updated July 2026, 71% of 799 first-time examinees passed the HVAC and Refrigeration exam, as did 40% of 229 repeat examinees (NCEES). Those figures describe groups of test takers, not your personal odds.
What score do I need to pass?
NCEES doesn't publish the passing score. Results are reported only as pass or fail (NCEES exam scoring).
Is there an official practice exam?
Yes. NCEES sells an 80-question HVAC and Refrigeration practice exam (PDF, $59.95 as of October 2026) based on the October 2025 specification. It includes alternative item types such as fill-in-the-blank and matching. It's available through MyNCEES exam prep, with a discount if you buy it when you register for the exam.
Do I need to pass the FE first, or have four years of experience?
Your state licensing board decides who can sit for the PE. NCEES designs the exam for engineers with at least four years of post-college experience, but some boards allow earlier testing. Check with your licensing board. If you still need the FE, start with FE Mechanical exam prep.
Does passing the exam make me a licensed PE?
No. Your licensing board issues the license, and passing the exam is one of its requirements.
How do registration, retakes, and accommodations work?
You register and pay through MyNCEES after checking your board's approval process. You can take the exam once per calendar-quarter testing window and no more than three times in 12 months, and some boards are stricter. Accommodations must be requested during registration. See NCEES exam registration, NCEES retake rules, NCEES exam results, and NCEES exam accommodations.
Want to compare HVAC with other PE exams? See PE exam prep.
Sources
Exam facts on this page were last verified October 8, 2026, against the NCEES sources below. We recalculated every practice problem and checked its method against the cited teaching source. The problems have not been reviewed by NCEES or by a licensed engineer. This guide was developed with AI-assisted research and editing; see how Castleport uses sources and AI assistance.
NCEES
- PE Mechanical exam page (format, fee, references, pass rates)
- PE Mechanical: HVAC and Refrigeration specification, effective October 2025
- PE Mechanical: HVAC and Refrigeration specification, effective April 2020 (historical)
- NCEES Examinee Guide, May 2026
- NCEES exam scoring
- NCEES notice of exam changes and 2026 calculator list
- NCEES HVAC and Refrigeration practice exam
Teaching sources for the practice problems
- ASHRAE Handbook online: Fundamentals (2025), Ch. 1 Psychrometrics; Fundamentals (2025), Ch. 2 Thermodynamics and Refrigeration Cycles; Fundamentals (2025), Ch. 8 Sound and Vibration; HVAC Systems and Equipment (2024), Ch. 40 Cooling Towers
- ASHRAE: Terminology; read-only standards list; Titles, Purposes, and Scopes
- NEBB Fundamental Formula Chart (2025)
- U.S. Department of Energy: Heat Transfer handbook, DOE-HDBK-1012/2-92; Fluid Flow handbook, DOE-HDBK-1012/3-92; Improving Pumping System Performance; Improving Fan System Performance; Improving Steam System Performance; Determining Electric Motor Load and Efficiency; FEMP chiller purchasing guide (2003)
- OSHA Technical Manual, Ventilation and Noise
- EPA Moisture Control Guidance
- PNNL-21569, building re-tuning measures
- NIST Handbook 135, Life Cycle Costing Manual
- UC Santa Barbara, Process Dynamics and Control, Ch. 8 slides; University of Iowa, vibration isolation lecture
These teaching sources explain the methods. They are not references supplied on the exam.
Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify their subjects; trademarks belong to their respective owners. The practice problems are original and unofficial, and nothing on this page guarantees a passing result or a license.
These practice explanations were prepared with AI assistance. They have not been reviewed by a named licensed mechanical engineer.